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I have a small telescope which shows me the moon's craters a little bit magnified. Consider a situation where the moon is very very brighter than its current brightness, so that the telescope's objective lens gathers much more light. Will the same telescope be able to show me the moon's craters in a more magnified image? If not, under what condition can this happen if the telescope is the same?

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There are two things to consider when it comes to magnification: what is the best resolution one can get, and will the image look good even if I have the resolution.

What is the best resolution one can get?

Light, as a wave, undergoes diffraction. Now, in the case of a simple telescope with a circular aperture1, an object with infinitesimal angular size will appear to have an angular size of2 $$ \theta_\text{best} = 1.22 \frac{\lambda}{D}, $$ where $\lambda$ is the wavelength light you are considering, $D$ is the telescope's diameter (in the same length units!), and $\theta_\text{best}$ is measured in radians. The system is then said to be diffraction-limited. The laws of physics won't let you disentangle features with angular sizes smaller than this.

So what happens if you keep trying for a higher magnification? You can magnify the image as much as you want, either with optics (discussed later) or with software. But it will end up more and more pixelated. Think of the diameter of your telescope as setting the limit on how many pixels you have. You can zoom in, but at some point the pixels will be too big and you'll lose quality.

As a side note, this is a hard limit for all systems. Other things limit $\theta_\text{best}$ in practice, the most notorious being the atmosphere. Any magnification by a factor of roughly a couple hundred or more will lead to an image that dances and blurs due to diffraction in the air. Minimizing this requires either (a) going up a very high mountain with a large ocean upwind of you or (b) very expensive corrective optics.

Will the image look good even if I have the resolution?

First we should be absolutely clear on how magnification works in telescopes. The magnification factor will be the ratio of the telescope's focal length (usually tens of centimeters to a couple of meters) to the eyepiece's focal length (usually from roughly $6$ to $40$ millimeters). That's it. Nothing else matters in setting this factor, such as the brightness of the object being observed.3

But let's say you have a nice, large-diameter telescope. The air is still. You put in a good eyepiece with a low focal length. The image is huge! And also black! What happened? Well, all that magnification spread the light a little too thinly. So in this sense you need a bright source to use a high-magnification setup. Looking at a dim galaxy under maximum magnification is a disappointing thing every amateur tries at least once before accepting that the best way to view such things is by studying the image harder, not trying to enlarge it too much.

On the other hand, the Moon is so enormously bright this is never an issue. Even using the highest magnification you can get your hands on, it will still be bright. If you have a small telescope (I assume that means a 4" diameter or less), you can probably get a more magnified image, and it won't be ruined by the effects mentioned above, but it requires a different eyepiece (assuming there are shorter focal lengths to be found).


1 Not that anyone uses square telescopes. The modification to this rule occur when you have objects blocking the aperture, so it is no longer a disc but maybe an annulus or something. Still, the modifications are small.

2 To be precise, that "point source" will appear as an Airy disk, where the $1.22$ is an approximation to the radius of the inner disk of that pattern.

3 This is why you should be wary of telescope vendors who taut their "high magnifications." Not only is high magnification not necessarily a good thing, it can be made whatever you want by swapping in a different eyepiece. (And pretty much all eyepieces come in one of two standard sizes, so they are really quite interchangeable.)

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No. The resolution you can resolve from a telescope depends on the wave-length and the width of the object. $$\theta=1.2\frac{\lambda }{d}$$

Try this:

http://en.wikipedia.org/wiki/Angular_resolution

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  • $\begingroup$ When considering the response of the retina, the brightness does effect what can be resolved. If the moon isn't illuminated you won't have much resolution. $\endgroup$ – Joe Oct 5 '13 at 22:49

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